$\int \left( \dfrac{x^3+2x^2-4}{x^2}\right)dx=$ $+C$
The integrand is the quotient of two functions: $x^3+2x^2-4$ and $x^2$. Although it is tempting to take the quotient of their integrals, this would not work. $\int \dfrac{f(x)}{g(x)}\,dx\neq\dfrac{\int f(x)\,dx}{\int g(x)\,dx}$ Instead, what we should do is divide each term in the numerator by the denominator. $\begin{aligned} &\phantom{=}\int \left( \dfrac{x^3+2x^2-4}{x^2}\right)dx \\\\ &=\int\left(\dfrac{x^3}{x^2}+\dfrac{2x^2}{x^2}-\dfrac{4}{x^2}\right)dx \\\\ &=\int (x+2-4x^{-2})\,dx \end{aligned}$ Now we can integrate using the reverse power rule, the sum rule, and the constant multiple rule for indefinite integrals. $\begin{aligned} &\phantom{=}\int \left( \dfrac{x^3+2x^2-4}{x^2}\right)dx \\\\ &=\int (x+2-4x^{-2})\,dx \\\\ &= \int x\,dx+2\int 1\,dx-4\int x^{-2}\,dx \\\\ &=\dfrac{x^2}{2}+2\dfrac{x^1}{1}-4\dfrac{x^{-1}}{-1}+C \\\\ &=\dfrac12x^2+2x+\dfrac{4}{x}+C \end{aligned}$ In conclusion, $\int \left( \dfrac{x^3+2x^2-4}{x^2}\right)dx=\dfrac12x^2+2x+\dfrac{4}{x}+C$